2D Integral, Gaussian and 2 Sinc Functions

[beautiful1]

I am looking for help with the following integral

$$A = \int dx \int dy \exp(-a (x+y)^2 +ib(x-y)) sinc(cx+dy) sinc(dx+cy)$$

where $$sinc(x) =\sin(x) / x$$ for $$x \neq 0$$ and $$sinc(0) = 1$$

(pls forgive my poor latex)

Either in the indefinite form or with the upper/lower limits at $$+/-\infty$$

The real-valued constants $$a, b, c,$$ and $$d$$ are positive.

My original idea was to switch to coordinates $$w = x+y$$ and $$u=x-y$$ but I can not get pass the sinc functions...any help would be appreciated.

 

[beautiful1]

I am looking for help with the following integral

$$A = \int dx \int dy \exp(-a (x+y)^2 +ib(x-y)) sinc(cx+dy) sinc(dx+cy)$$

I have now transformed coordinates using

$$w = x + y$$

$$u = x - y$$

$$dw du = -2 dx dy$$

and I get

$$A = -\frac{1}{2} \int dw e^{-a w^2} \int du e^{ibu} {\rm sinc}(\alpha w + \beta u) {\rm sinc}(\alpha w - \beta u)$$

where I define

$$\alpha = (c+d)/2$$ and $$\beta = (c-d)/2$$

Noting that

$${\rm sinc}[\alpha w + \beta u] {\rm sinc}[\alpha w - \beta u] = -\frac{1}{2} \frac{(\cos[2\alpha w]-\cos[2\beta u])}{(\alpha w)^2 - (\beta u)^2}$$

I find

$$4A= \int dw e^{-a w^2} \cos[2\alpha x] \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} -\int dw e^{-a w^2} \int du e^{i \beta u} \frac{\cos[2\beta u]}{(\alpha w)^2 - (\beta u)^2}$$

I do not know what to do from here.

I have put each of the integrals over u into "the integrator" at http://integrals.wolfram.com/, which returns an answer in terms of the exponential integral $$Ei(x)$$, but I am unsure how to get to that point. I have not tried to perform the subsequent integral over w.

Any help would be appreciated.

 

[beautiful1]

In case anyone cares

I find

$$4A= \int dw e^{-a w^2} \cos[2\alpha x] \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} -\int dw e^{-a w^2} \int du e^{i \beta u} \frac{\cos[2\beta u]}{(\alpha w)^2 - (\beta u)^2}$$

For the first integral over u

$$\begin{equation*} \begin{split} I_{1}(x) &= \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} \\ \\ & = e^{-i \alpha w} \int du \frac{e^{i \beta u}}{(\beta u)(\alpha w - \beta u)} \\ \\ & = \frac{e^{-i \alpha w} }{2\alpha w} \int du e^{i \beta u} \left( \frac{1}{\beta u} + \frac{1}{\alpha w + \beta u} \right) \end{split} \end{equation*}$$

My interest is in the case the upper and lower limits of integration are + and - $$\infty$$, respectively.

Then

$$\begin{equation*} \begin{split} \int_{-\infty}^{\infty} du \frac{e^{iu}}{u} &= \int _{0}^{\infty} du \frac{e^{iu}}{u} + \int _{-\infty}^{0} du \frac{e^{iu}}{u} \\ \\ & = \int _{0}^{\infty} du \frac{e^{iu}}{u} - \int _{0}^{\infty} du \frac{e^{-iu}}{u} \\ \\ &= \frac{1}{2i} \int _{0}^{\infty} du \frac{\sin u}{u} \end{split} \end{equation*}$$

The last integral is the sine integral. Note the integrand is the sinc function. And (somehow, not sure yet, may be another post) for the given limits, this evaluates to

$$\int _{0}^{\infty} du \frac{\sin u}{u} = \frac{\pi}{2}$$

I believe the remaining integrals over u will all evaluate in a similar manner.
As for w...